In the reaction :
`CH_(3)-C-=C-Hoverset(NaNH_(2)//NH_(3)(l))rarr(A)overset(CH_(3)-overset(Br)overset(|)underset(CH_(3))underset("| ")"C "-CH_(3))rarr(B)`,
The product (B) is :

Options

A. Only `CH_(3)-C-=overset(CH_(3))overset("| ")underset(CH_(3))underset("| ")"C "-C-CH_(3)`
B. Only `CH_(3)-underset(CH_(3))underset("| ")"C "-CH_(2)`
C. `1:1" mixture of "CH_(3)-C-=overset(CH_(3))overset("| ")underset(CH_(3))underset("| ")"C "-C-CH_(3) and CH_(3)-underset(CH_(3))underset("| ")"C "=CH_(2)`
D. `1:1" mixture of "CH_(3)-underset(CH_(3))underset("| ")"C "=CH_(2) and CH_(3)-C-=C-H`

Explanation

To solve the given reaction step-by-step, let's break it down:

### Step 1: Identify the starting material
The starting material is **CH₃-C≡C-H** (1-butyne).

### Step 2: Reaction with NaNH₂ in liquid ammonia
When 1-butyne reacts with **NaNH₂** (sodium amide) in liquid ammonia, NaNH₂ acts as a strong base. It abstracts a proton (H⁺) from the terminal alkyne (the hydrogen attached to the carbon with the triple bond), resulting in the formation of a **carbanion**.

**Reaction:**
\[ \text{CH}_3-C \equiv C-H + \text{NaNH}_2 \rightarrow \text{CH}_3-C \equiv C^- + \text{NH}_3 \]

This carbanion is denoted as **A**:
\[ \text{A} = \text{CH}_3-C \equiv C^- \]

### Step 3: Reaction with a tertiary alkyl halide
Next, the carbanion (A) reacts with a tertiary alkyl halide, which is **(CH₃)₃CBr** (tert-butyl bromide). The carbanion will act as a nucleophile and attack the carbon atom of the tertiary halide, leading to the formation of a new carbon-carbon bond.

**Reaction:**
\[ \text{CH}_3-C \equiv C^- + \text{(CH}_3)_3CBr \rightarrow \text{CH}_3-C \equiv C-(C(CH_3)_3) + \text{Br}^- \]

### Step 4: Identify the product B
The product formed from this reaction is a compound where the carbanion has added to the tertiary carbon, resulting in a new alkyne. The product (B) can be represented as:
\[ \text{B} = \text{CH}_3-C \equiv C-C(CH_3)_3 \]

### Step 5: Final structure of product B
The final structure of product B is:
\[ \text{B} = \text{(CH}_3)_2C=C(CH_3)C \]

### Conclusion
Thus, the product (B) formed from the reaction is a substituted alkyne.